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mojhsa [17]
2 years ago
9

❊ Simplify :

Mathematics
2 answers:
Rzqust [24]2 years ago
8 0

Answer:

<em>Your</em><em> </em><em>solution</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

Nataliya [291]2 years ago
5 0
<h3>Need to Do :- </h3>
  • To simplify the given expression .

\red{\frak{Given}}\Bigg\{ \sf \dfrac{x - 1}{ {x}^{2} - 3x + 2} + \dfrac{x - 2}{ {x}^{2} - 5x + 6 } + \dfrac{x - 5}{ {x}^{2} - 8x + 15 }

\rule{200}4

\sf\longrightarrow \small  \dfrac{x - 1}{ {x}^{2} - 3x + 2} + \dfrac{x - 2}{ {x}^{2} - 5x + 6 } + \dfrac{x - 5}{ {x}^{2} - 8x + 15 } \\\\\\\sf\longrightarrow \small  \dfrac{ x-1}{x^2-x -2x +2} +\dfrac{ x-2}{x^2-3x-2x+6} +\dfrac{ x -5}{x^2-5x -3x + 15 } \\\\\\\sf\longrightarrow\small \dfrac{ x -1}{ x ( x - 1) -2(x-1) } +\dfrac{ x-2}{x ( x -3) -2( x -3)} +\dfrac{ x -5}{ x(x-5) -3( x -5) }  \\\\\\\sf\longrightarrow \small \dfrac{ x -1}{ ( x-2) (x-1) } +\dfrac{ x-2}{( x -2)(x-3) } +\dfrac{ x -5}{ (x-3)(x-5)  } \\\\\\\sf\longrightarrow\small \dfrac{ 1}{ x-2} +\dfrac{ 1}{ x -3} +\dfrac{1}{ x -3}   \\\\\\\sf\longrightarrow   \small  \dfrac{1}{x-2} +\dfrac{2}{x-3}  \\\\\\\sf\longrightarrow   \small \dfrac{ x-3 +2(x-2)}{ ( x -3)(x-2) }  \\\\\\\sf\longrightarrow   \small \dfrac{ x - 3 +2x -4 }{ (x-3)(x-2) }     \\\\\\\sf\longrightarrow   \underset{\blue{\sf Required \ Answer  }}{\underbrace{\boxed{\pink{\frak{  \dfrac{ 3x -7}{ ( x -2)(x-3) } }}}}}

\rule{200}4

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Colin has a pad with x pieces of paper on it. For his first class, he wrote on 5 fewer than half of the pieces of paper in the p
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Answer:

Colin has <em>8 sheets </em>left for his third class.

Step-by-step explanation:

Given that:

Total Number of pieces of papers = x

Number of pieces of papers used for 1st class = 5 fewer than half of the pieces in the pad

Writing the equation:

\text{Number of pieces of papers used for 1st class =} \dfrac{x}{2} -5 ...... (1)

Also, Given that number of pieces of papers used for the 2nd class are 2 more than that of papers used in the 1st class.

\text{Number of pieces of papers used for 2nd class =} \dfrac{x}{2} -5+2 = \dfrac{x}2 -3 ...... (2)

Now, number of pieces of papers left for the third class = Total number of pieces of papers in the pad - Number of pieces of papers used in the first class - Number of pieces of papers used in the first class

\text{number of pieces of papers left for the third class = }x-(\dfrac{x}{2}-5)-(\dfrac{x}{2}-3)\\\Rightarrow x-\dfrac{x}2-\dfrac{x}2+5+3\\\Rightarrow x-x+5+3\\\Rightarrow 8

So, the answer is:

Colin has <em>8</em> <em>sheets </em>left for his third class.

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

2 or -2

Step-by-step explanation:

If x² - 8x = -12

Step 1

We find x by solving

x² - 8x = -12

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x² - 8x + 12 = 0

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